Qualitative Analysis - Result Question 24
####24. The correct option(s) to distinguish nitrate salts to $Mn^{2+}$ and $Cu^{2+}$ taken separately is (are)
(2018 Adv.)
(a) $Mn^{2+}$ shows the characteristic green colour in the flame test
(b) Only $Cu^{2+}$ shows the formation of precipitate by passing $H _2 S$ in acidic medium
(c) Only $Mn^{2+}$ shows the formation of precipitate by passing $H _2 S$ in faintly basic medium
(d) $Cu^{2+} / Cu$ has higher reduction potential than $Mn^{2+} / Mn$ (measured under similar conditions)
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Answer:
Correct Answer: 24. (b,d)
Solution:
- Statement wise explanation is
Statement (a) $Mn^{2+}$ produces yellow-green colour in flame test while $Cu^{2+}$ produces bluish-green colour in flame test. Thus, due to the presence of green colour in both the cases, flame test is not the suitable method to distinguish between nitrate salts of $Cu^{2+}$ and $Mn^{2+}$. Hence this statement is wrong.
Statement (b) $Cu^{2+}$ belong to group II of cationic or basic radicals. It gives black ppt. of $CuS$ if $H _2 S$ is passed through it in the presence of acid (e.g HCl). $Mn^{2+}$ does not show this property hence this can be considered as a suitable method to distinguish between $Mn^{2+}$ and $Cu^{2+}$.
Hence, this statement is correct.
Statement (c) In faintly basic medium when $H _2 S$ is passed both $Cu^{2+}$ and $Mn^{2+}$ forms precipitates. Thus, it is not suitable method to distinguish between them.
Hence, this statement is incorrect
Statement (d) The standard reduction potential of $Cu^{2+} / Cu$ is $+0.34 V$ while that of $Mn^{2+} / Mn$ is $-1.18 V$. This can be used to distinguish between $Cu^{2+}$ and $Mn^{2+}$. In general less electropositive metals have higher SRP.
Hence, this statement is correct.
